Steady Viscometric Flows

Abstract

There is a class of flows of the simple fluid, equivalent to the simple shearing flow, for which the kinematics and the stress can be completely determined. (Ericksen, Viscoelasticity - Phenomenological Aspects, Academic Press, New York 1960) Ericksen [21] called them laminar shear flows, but the current term used to describe these flows is viscometric flows (Coleman, Arch. Rat. Mech. Anal. 9, 273–300, 1962) [16]. We review this class of flows here.

Problems

Problem 6.1

Show that the velocity gradient for (6.3) is

$$\begin{aligned} \mathbf {L} = \dot{\gamma }\mathbf {ab}. \end{aligned}$$

(6.22)

Consequently, show that all the flows represented by (6.3) are isochoric. Show that the shear rate is $\left| {\dot{\gamma }} \right| .$

Problem 6.2

Show that the shear rate for the helicoidal flow (6.14) is

$$\begin{aligned} \dot{\gamma }^{2}=\left( {r^{2}+c^{2}}\right) \nabla \omega \cdot \nabla \omega . \end{aligned}$$

(6.23)

Problem 6.3

Consider the shear flow between two tilted plates: the first plate is at rest and the second plate, tilted at an angle $\theta _{0}$ to the first plate, is moving with a velocity U in the $\mathbf {k}-$direction, as shown in Fig. 6.2. Show that

$$\begin{aligned} \mathbf {u}=U\frac{\theta }{{\theta _{0}}}\mathbf {e}_{z}. \end{aligned}$$

(6.24)

Show that the stress is given by

$$\begin{aligned} \mathbf {T}=-P\mathbf {I}+\eta \dot{\gamma }\left( {\mathbf {e}_{z}\mathbf {e} _{\theta }+\mathbf {e}_{\theta }\mathbf {e}_{z}}\right) +\left( {N_{1}+N_{2}} \right) \mathbf {e}_{z}\mathbf {e}_{z}+N_{2}\mathbf {e}_{\theta }\mathbf {e} _{\theta }, \end{aligned}$$

(6.25)

where the shear rate is $\dot{\gamma }=U/r\theta _{0},$ and

$$\begin{aligned} P=P\left( {r_{0}}\right) +I_{2}\left( {\dot{\gamma }}\right) -I_{2}\left( { \dot{\gamma }_{0}}\right) ,\;\;\;I_{2}\left( {\dot{\gamma }}\right) =\int _{0}^{ \dot{\gamma }}{\dot{\gamma }\nu _{2}d\dot{\gamma }}. \end{aligned}$$

(6.26)

Suggest a way to measure $N_{2}$ based on this.

Problem 6.4

The flow between two parallel, coaxial disks is called torsional flow. In this flow, the bottom disk is fixed, and the top disk rotates at an angular velocity of $\varOmega$. The distance between the disks is h. Neglecting the fluid inertia, show that

$$\begin{aligned} \mathbf {u}=\varOmega r \frac{z}{h}\mathbf {e}_{\theta },\;\;\;\dot{\gamma }=\varOmega \frac{r}{h}. \end{aligned}$$

(6.27)

Show that the torque required to turn the top disk is

$$\begin{aligned} M=2\pi \int _{0}^{R}{\dot{\gamma }\eta \left( {\dot{\gamma }}\right) r^{2}dr}, \end{aligned}$$

(6.28)

where R is the radius of the disks. Show that the pressure is

$$\begin{aligned} P\left( r\right) =\int _{\dot{\gamma }}^{\dot{\gamma }_{R}}{\dot{\gamma }\left( { \nu _{1}+\nu _{2}}\right) d\dot{\gamma }}. \end{aligned}$$

(6.29)

From the axial stress, show that the normal force on the top disk is

$$\begin{aligned} F=\pi R^{2}\dot{\gamma }_{R}^{-2}\int _{0}^{\gamma _{R}}{\dot{\gamma }\left( { N_{1}-N_{2}}\right) d\dot{\gamma }}, \end{aligned}$$

(6.30)

where $\dot{\gamma }_{R}=\varOmega R/h$ is the shear rate at the rim $r=R.$

By normalizing the torque and the force as

$$\begin{aligned} m=\frac{M}{{2\pi R^{3}}},\;\;\;f=\frac{F}{{\pi R^{2}}}, \end{aligned}$$

(6.31)

show that

$$\begin{aligned} \eta \left( {\dot{\gamma }_{R}}\right) =\frac{m}{{\dot{\gamma }_{R}}}\left[ {3+ \frac{{d\ln m}}{{d\ln \dot{\gamma }_{R}}}}\right] , \end{aligned}$$

(6.32)

and

$$\begin{aligned} N_{1}\left( {\dot{\gamma }_{R}}\right) -N_{2}\left( {\dot{\gamma }_{R}}\right) =f\left( {2+\frac{{d\ln f}}{{d\ln \dot{\gamma }_{R}}}}\right) . \end{aligned}$$

(6.33)

Relations (6.32), (6.33) are the basis for the operation of the parallel-disk viscometer.

Problem 6.5

In a pipe flow, of radius R and pressure drop/unit length $\varDelta P/L$, show that the flow rate is

$$\begin{aligned} Q=8\pi \left( { \frac{L}{{\varDelta P}}}\right) ^{3}\int _{0}^{\tau _{w}}{\frac{{\tau ^{3}}}{\eta } d\tau }, \end{aligned}$$

(6.34)

where $\tau$ is the shear stress, and $\tau _{w}$ is the shear stress at the wall. In terms of the reduced discharge rate,

$$\begin{aligned} q=\frac{Q}{{\pi R^{3}}}, \end{aligned}$$

(6.35)

show that

$$\begin{aligned} \frac{{dq}}{{d\tau _{w}}}=\frac{1}{{\eta \left( {\tau _{w}}\right) }}-\frac{{3q} }{{\tau _{w}}}, \end{aligned}$$

(6.36)

or

$$\begin{aligned} \dot{\gamma }_{w}=q\left( {\tau _{w}}\right) \left[ {3+\frac{{d\ln q}}{{ d\ln \tau _{w}}}}\right] . \end{aligned}$$

(6.37)

The relation (6.37) is due to Rabinowitch [74] and is the basis for capillary viscometry.